Tóm tắt :
\(l_1=100m\)
\(S_1=1mm^2=1.10^{-6}m^2\)
R1 = 1,7 Ω
\(S_2=0,2mm^2=0,2.10^{-6}m^2\)
\(R_2=17\Omega\)
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\(l_2=?\)
GIẢI :
Ta có : \(\dfrac{R_1}{R_2}=\dfrac{\rho.\dfrac{l_1}{S_1}}{\rho.\dfrac{l_2}{S_2}}=\dfrac{\dfrac{100}{10^{-6}}}{\dfrac{l_2}{0,2.10^{-6}}}=\dfrac{100}{10^{-6}}.\dfrac{0,2.10^{-6}}{l_2}=\dfrac{20}{l_2}\)
=> \(\dfrac{1,7}{17}=\dfrac{20}{l_2}\)
\(\rightarrow l_2=\dfrac{17.20}{1,7}=200\left(m\right)\)
Vậy dây đồng thứ 2 có chiều dài là 200m